Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int C = 300001;
struct point {
  double x, y;
};
struct seg {
  point a, b;
};
struct line {
  double a, b;
};
double sy, l[C], r[C], sum[C], x, y;
int n, i, j, q;
point p, beg, end;
line k;
double bx, ex, z;
line fun(point a, point b) {
  line g;
  g.a = (a.y - b.y) / (a.x - b.x);
  g.b = a.y - (g.a * a.x);
  return g;
}
inline int binup(double tab[], double v, int l, int r) {
  int m;
  while (l <= r) {
    m = (l + r) / 2;
    if (tab[m] >= v)
      r = m - 1;
    else
      l = m + 1;
  }
  return l;
}
inline int binlow(double tab[], double v, int l, int r) {
  int m;
  while (l <= r) {
    m = (l + r) / 2;
    if (tab[m] > v)
      r = m - 1;
    else
      l = m + 1;
  }
  return r;
}
double sear(double a, double b) {
  int f, g;
  double s = 0;
  f = binup(r, a, 0, n - 1);
  g = binlow(l, b, 0, n - 1);
  if (g == f) {
    if (b > r[f] && a < l[f])
      s = s + r[f] - l[f];
    else if (b > r[f])
      s = s + r[f] - a;
    else if (a < l[f])
      s = s + b - l[f];
    else
      s = b - a;
  } else if (g > f) {
    s = s + sum[g - 1] - sum[f];
    if (b > r[g])
      s = s + r[g] - l[g];
    else
      s = s + b - l[g];
    if (a < l[f])
      s = s + r[f] - l[f];
    else
      s = s + r[f] - a;
  }
  return s / (b - a);
}
int main() {
  scanf("%lf %lf %lf", &beg.y, &beg.x, &end.x), end.y = beg.y;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%lf %lf", &l[i], &r[i]);
    if (i > 0)
      sum[i] = r[i] - l[i] + sum[i - 1];
    else
      sum[i] = r[i] - l[i];
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%lf %lf", &p.x, &p.y);
    if (p.x == beg.x)
      bx = p.x;
    else
      k = fun(p, beg), bx = -k.b / k.a;
    if (p.x == end.x)
      ex = p.x;
    else
      k = fun(p, end), ex = -k.b / k.a;
    z = sear(bx, ex);
    printf("%.10lf\n", z * (end.x - beg.x));
  }
  return 0;
}